Predict the product of the following rearrangement-prone E1 el...

Question

Predict the product of the following rearrangement-prone E1 eliminations.

(b)

Accepted Answer

Hi, everybody. Welcome back. Our next question says, what is the product of the given E one reaction that is prone to rearrangement? And we have a reagent that shows a cyclops ring with a Proppy chain that has a bromme on carbon one, it's reacting in ethanol with heat. So we're already told we'll have an E one reaction. So we'll have an elimination reaction in two steps. So we know that the leaving group we'll leave first as our first step and that, that will produce a Carbocaine intermediate. And we're already told that our Carbocaine intermediate is prone to rearrangement. So we know what to look for here. So let's look at our reagent and think about our first step, which would be our leaving group going away. When we look at this molecule, it's pretty obvious to see that bromine is going to be the obvious choice of a leaving group. So we can show the electrons from this carbon bromine bond going to the bromine and we're left with our initial Carbocaine. So we have our cyclops ring and then it's purple chain, but now where the bromine was, we have a positive charge. So there's our initial Carbocaine, it's a secondary carbo CDI. So we should be looking around again, especially because we've got this in given us for a more stable carrot so that the, we can see we'll have this rearrangement to lead to a more stable carbo caravan. So what would be more stable and secondary while a tertiary Carbocaine, which we actually could obtain by a hydride shift. We have a hydrogen here on this first carbon in the ring. So we could shift that to the positively charged carbon and have a tertiary Carbocaine. But we have something even better than creating a tertiary carbo cion because we have our positive charts adjacent 285 carbon ring. And when that positive charge is adjacent to a 34 or five carbon ring, we therefore have the possibility of ring expansion. So while a tertiary Carbocaine is indeed more stable than a secondary, we can have greater stability of the entire molecule by expanding that ring and releasing some of the ring strain due to the smaller pentane size. So we can have our next step here our ring expansion. So for me to keep track of a ring expansion, I like to use different colored dots in the carbons. If you want to test and don't have colored pencils or markers, you could number the carbons to keep track. But I find it a little less cluttered to use the colored dots. So I'm going to use green on the carbon that has the positive charge and then blue on that first carbon in the ring that is adjacent to our marine carbon. And then for the carbon after that, you see our ring as symmetrical, it doesn't matter which direction I go, it will lead to the same result. So I'm just going to the left here and showing the next carbon in the ring as a red carbon. So we have green on the chain blue on the ring and then red adjacent to it on the ring. So let's think about how this ring will expand. Well, the electrons that form the bond between the blue carbon and the red carbon. We'll come on over to the positively charged carbon that will break the bond between the blue and red, opening up the ring and create a new bond between the red and green carbon. So we've added another carbon to the ring. So our product will be a cyclohexane ring which again more stable than cyclops. Its angles in the ring are pretty much the same as the preferred tetrahedral angles. So much more favorable, set up for those carbon, the carbons in the ring. And now let's keep track of our carbons here. So our green carbon is now within the ring. So I'm going to label the bottom carbon on the ring with a green dot It's been incorporated there by its new bond to the red carpet. So the car, the carbon to the left and the ring, I'm going to leave with a red dot because they are now within the ring adjacent to each other. The bond between the red and blue carbon has been broken. But note the blue carbon of course, is still connected to the green carbon from when it was in the chain. So that blue carbon is now on the other side of the green carbon within the ring. So they were in the order red, blue, green, they're now in the order red, green, blue. So you can see why it can be a little tricky to keep straight which carbon is, which which is important if you have substituents on them to keep track of where they're going. So our green carbon was part of a chain, was part of a three carbon chain. So there's still two more carbons attached to it. So we need to add those two carbons in the form of an ethyl group onto the green carbon. Now, where did our positive charge go? Well, let's think of the hydrogens on these carbons. So our red carbon had no substitutes and was in a ring. So it had two hydrogens. Our blue carbon was a tertiary carbon with three bonds to the carbon. So it had one hydrogen and our green carbon there with its positive charge and two bonds to a carbon had one hydrogen. That's not the proper geometry there. It should be trigonal, but my positive charge is in the way. So just indicating that the hydrogen is there. So let's think about what happens after the ring expansion. Our red carbon within the ring, no substituent still has two hydrogens. Our green carbon has just one hydrogen, but it's now in this position where it has three bonds to other carbons, it only needs one hydrogen. So it's now neutral instead of positive. But our blue carbon only had one hydrogen. It's now within this ring with no substituents. So our blue carbon now has the positive charge because it has only three bonds. So that's why it's a Carbocaine rearrangement. We still have a Carbocaine, but it's been rearranged in terms of where that positive charge is. And again, this is more stable due to expanded ring size. So this is the pathway we'll go through to produce our major product since again, that more stable Carbocaine will make it the favored pathway. So now what happens here, we are going to deproteinate. So, dep protonation of the beta hydrogen. So we have two possibilities as we have two beta carbons here going to label them in red. We've got a beta carbon further up in the ring on one of these carbons with no substituents. And we have a beta carbon. Our green carbon is now a beta carbon to our carbo. Well, how do we decide which will give us the major product? The one that gives the more substitute alkene, that will be the more stable alkene and therefore the favored pathway. So we're going to deproteinate at the green carbon to produce that more substitute alkene. So I'm going to draw my ethanol in the form of Hoet put those two lone pairs on the oxygen. So the electrons from one of those lump pairs in the oxygen will grab that beta hydrogen, which I better draw on here. So let's draw hydrogen on our green carbon. And these electrons come over here and grab it in the form of a proton. And then the electrons from that carbon hydrogen bond come down to form a double bond between our blue and green carbons. So at least our major product here got our cyclohexane ring, an ethel group and now we formed a double bond. So between this carbon that has the ethyl group and the carbon adjacent to it. So there we go, we have our major product here. Now important to note that you might get some percent of perhaps alkanes resulting from the original Carbocaine on either side or an alkane resulting from the dep protonation of the other beta hydrogen. But the major product will be the one I have drawn because it's the product that results from the most stable Carbocaine and then gives the most stable alkane. So this is the major product of this reaction. So that's how we will label that. So see you in the next video

Predict the product of the following rearrangement-prone E1 eliminations.
(b)

Key Concepts

E1 Mechanism

Carbocation Stability

Rearrangement in Organic Reactions

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